The Stacks project

Lemma 29.49.1. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $\eta \in Y$ be a generic point of an irreducible component of $Y$. The following are equivalent:

  1. the set $f^{-1}(\{ \eta \} )$ is finite,

  2. there exist affine opens $U_ i \subset X$, $i = 1, \ldots , n$ and $V \subset Y$ with $f(U_ i) \subset V$, $\eta \in V$ and $f^{-1}(\{ \eta \} ) \subset \bigcup U_ i$ such that each $f|_{U_ i} : U_ i \to V$ is finite.

If $f$ is quasi-separated, then these are also equivalent to

  1. there exist affine opens $V \subset Y$, and $U \subset X$ with $f(U) \subset V$, $\eta \in V$ and $f^{-1}(\{ \eta \} ) \subset U$ such that $f|_ U : U \to V$ is finite.

If $f$ is quasi-compact and quasi-separated, then these are also equivalent to

  1. there exists an affine open $V \subset Y$, $\eta \in V$ such that $f^{-1}(V) \to V$ is finite.

Proof. The question is local on the base. Hence we may replace $Y$ by an affine neighbourhood of $\eta $, and we may and do assume throughout the proof below that $Y$ is affine, say $Y = \mathop{\mathrm{Spec}}(R)$.

It is clear that (2) implies (1). Assume that $f^{-1}(\{ \eta \} ) = \{ \xi _1, \ldots , \xi _ n\} $ is finite. Choose affine opens $U_ i \subset X$ with $\xi _ i \in U_ i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ each of the morphisms $U_ i \to Y$ is finite. In other words (2) holds.

It is clear that (3) implies (1). Assume $f^{-1}(\{ \eta \} ) = \{ \xi _1, \ldots , \xi _ n\} $ and assume that $f$ is quasi-separated. Since $Y$ is affine this implies that $X$ is quasi-separated. Since each $\xi _ i$ maps to a generic point of an irreducible component of $Y$, we see that each $\xi _ i$ is a generic point of an irreducible component of $X$. By Properties, Lemma 28.29.1 we can find an affine open $U \subset X$ containing each $\xi _ i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ the morphisms $U \to Y$ is finite. In other words (3) holds.

It is clear that (4) implies all of (1) – (3) with no further assumptions on $f$. Suppose that $f$ is quasi-compact and quasi-separated. We have to show that the equivalent conditions (1) – (3) imply (4). Let $U$, $V$ be as in (3). Replace $Y$ by $V$. Since $f$ is quasi-compact and $Y$ is quasi-compact (being affine) we see that $X$ is quasi-compact. Hence $Z = X \setminus U$ is quasi-compact, hence the morphism $f|_ Z : Z \to Y$ is quasi-compact. By construction of $Z$ we see that $\eta \not\in f(Z)$. Hence by Lemma 29.8.4 we see that there exists an affine open neighbourhood $V'$ of $\eta $ in $Y$ such that $f^{-1}(V') \cap Z = \emptyset $. Then we have $f^{-1}(V') \subset U$ and this means that $f^{-1}(V') \to V'$ is finite. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02NW. Beware of the difference between the letter 'O' and the digit '0'.